23. Group Actions and Automorphisms Recall the Definition of an Action

Home , Automorphism, Automorphism group, Coset, Homomorphism, Inner automorphism, Outer automorphism group

23. Group actions and automorphisms Recall the definition of an action: Definition 23.1. Let G be a group and let S be a set. An action of G on S is a function G × S −→ S denoted by (g, s) −→ g · s, such that e · s = s and (gh) · s = g · (h · s) In fact, an action of G on a set S is equivalent to a group homomorphism (invariably called a representation) ρ: G −→ A(S). Given an action G × S −→ S, define a group homomorphism ρ: G −→ A(S) by the rule ρ(g) = σ : S −→ S, where σ(s) = g · s. Vice-versa, given a representation (that is, a group homomorphism) ρ: G −→ A(S), define an action G · S −→ S by the rule g · s = ρ(g)(s). It is left as an exercise for the reader to check all of the details. The only sensible way to understand any group is let it act on some- thing. Definition-Lemma 23.2. Suppose the group G acts on the set S. Define an equivalence relation ∼ on S by the rule s ∼ t if and only if g · s = t for some g ∈ G. The equivalence classes of this action are called orbits. The action is said to be transitive if there is only one orbit (neces- sarily the whole of S). Proof. Given s ∈ S note that e · s = s, so that s ∼ s and ∼ is reflexive. If s and t ∈ S and s ∼ t then we may find g ∈ G such that t = g · s. But then s = g−1 · t so that t ∼ s and ∼ is symmetric. If r, s and t ∈ S and r ∼ s, s ∼ t then we may find g and h ∈ G such that s = g · r and t = h · s. In this case t = h · s = h · (g · r) = (hg) · r, so that t ∼ r and ∼ is transitive. 1 Definition-Lemma 23.3. Suppose the group G acts on the set S. Given s ∈ S the subset H = { g ∈ G | g · s = s }, is called the stabiliser of s ∈ S. H is a subgroup of G. Proof. H is non-empty as it contains the identity. Suppose that g and h ∈ H. Then (gh) · s = g · (h · s) = g · s = s. Thus gh ∈ H, H is closed under multiplication and so H is a subgroup of G. Example 23.4. Let G be a group and let H be a subgroup. Let S be the set of all left cosets of H in G. Define an action of G on S, G × S −→ S as follows. Given gH ∈ S and g0 ∈ G, set g0 · (gH) = (g0g)H. It is easy to check that this action is well-defined. Clearly there is only one orbit and the stabiliser of the trivial left coset H is H itself. Lemma 23.5. Let G be a group acting transitively on a set S and let H be the stabiliser of a point s ∈ S. Let L be the set of left cosets of H in G. Then there is an isomorphism of actions (where isomorphism is defined in the obvious way) of G acting on S and G acting on L, as in (23.4). In particular |G| |S| = . |H| Proof. Define a map f : L −→ S by sending the left coset gH to the element g ·s. We first have to check that f is well-defined. Suppose that gH = g0H. Then g0 = gh, for some h ∈ H. But then g0 · s = (gh) · s = g · (h · s) = g · s. Thus f is indeed well-defined. f is clearly surjective as the action of G is transitive. Suppose that f(gH) = f(gH). Then gS = g0s. In this case h = g−1g0 stabilises s, so that g−1g0 ∈ H. But then g and g0 are 2 in the same left coset and gH = g0H. Thus f is injective as well as surjective, and the result follows. Given a group G and an element g ∈ G recall the centraliser of g in G is Cg = { h ∈ G | hg = gh }. The centre of G is then Z(G) = { h ∈ H | gh = hg }, the set of elements which commute with everything; the centre is the intersection of the centralisers. Lemma 23.6 (The class equation). Let G be a group. The cardinality of the conjugacy class containing g ∈ G is the index of the centraliser, [G : Cg]. Further X |G| = |Z(G)| + [G : Cg],

[G:Cg]>1 where the second sum run over those conjugacy classes with more than one element. Proof. Let G act on itself by conjugation. Then the orbits are the conjugacy classes. If g ∈ then the stabiliser of g is nothing more than the centraliser. Thus the cardinality of the conjugacy class containing g is [G : Cg] by (23.3). If g ∈ G is in the centre of G then the conjugacy class containing G has only one element, and vice-versa. As G is a disjoint union of its conjugacy classes, we get the second equation. Lemma 23.7. If G is a p-group then the centre of G is a non-trivial subgroup of G. In particular G is simple if and only if the order of G is p. Proof. Consider the class equation X |G| = |Z(G)| + [G : Cg].

[G:Cg]>1 The ﬁrst and last terms are divisible by p and so the order of the centre of G is divisible by p. In particular the centre is a non-trivial subgroup. If G is not abelian then the centre is a proper normal subgroup and G is not simple. If G is abelian then G is simple if and only if its order is p. Theorem 23.8. Let G be a ﬁnite group whose order is divisible by a prime p. Then G contains at least one Sylow p-subgroup. 3 Proof. Suppose that n = pkm, where m is coprime to p. Let S be the set of subsets of G of cardinality pk. Then the cardinality of S is given by a binomial

n pkm(pkm − 1)(pkm − 2) ... (pkm − pk + 1) = pk pk(pk − 1) ... 1 Note that for every term in the numerator that is divisible by a power of p, we can match this term in the denominator which is also divisible by the same power of p. In particular the cardinality of S is coprime to p. Now let G act on S by left translation,

G × S −→ S where (g, P ) −→ gP.

Then S is breaks up into orbits. As the cardinality is coprime to p, it follows that there is an orbit whose cardinality is coprime to p. Suppose that X belongs to this orbit. Pick g ∈ X and let P = g−1X. Then P contains the identity. Let H be the stabiliser of P . Then H ⊂ P , since h · e ∈ P . On the other hand, [G : H] is coprime to p, so that the order of H is divisible by pk. It follows that H = P . But then P is a Sylow p-subgroup.

Question 23.9. What is the automorphism group of Sn? Deﬁnition-Lemma 23.10. Let G be a group. If a ∈ G then conjugation by G is an automorphism σa of G, called an inner automorphism of G. The group G0 of all inner automorphisms is isomorphic to G/Z, where Z is the centre. G0 is a normal subgroup of Aut(G) the group of all automorphisms and the quotient is called the outer automorphism group of G.

Proof. There is a natural map

ρ: G −→ Aut(G), whose image is G0. The kernel is isomorphic to the centre and so

G0 ' G/Z, by the ﬁrst Isomorphism theorem. It follows that G0 ⊂ Aut(G) is a subgroup. Suppose that φ: G −→ G is any automorphism of G.I claim that −1 φσaφ = σφ(a). 4 Since both sides are functions from G to G it suﬃces to check they do the same thing to any element g ∈ G. −1 −1 −1 φσaφ (g) = φ(aφ (g)a ) = φ(a)gφ(a)−1

= σφ(a)(g). 0 Thus G is normal in Aut(G).

Lemma 23.11. The centre of Sn is trivial unless n = 2. Proof. Easy check.

Theorem 23.12. The outer automorphism group of Sn is trivial unless n = 6 when it is isomorphic to Z2.

Lemma 23.13. If φ: Sn −→ Sn is an automorphism of Sn which sends a transposition to a transposition then φ is an inner automorphism. Proof. Since any automorphism permutes the conjugacy classes, φ sends transpositions to transpositions. Suppose that φ(1, 2) = (i, j). Let a = (1, i)(2, j). Then σa(i, j) = (1, 2) and so σaφ fixes (1, 2). It is ob- viously enough to show that σaφ is an inner automorphism. Replacing φ by σaφ we may assume φ fixes (1, 2). Now consider τ = φ(2, 3). By assumption τ is a transposition. Since (1, 2) and (2, 3) both move 2, τ must either move 1 or 2. Suppose it moves 1. Let a = (1, 2). Then σaφ still fixes (1, 2) and σaτ moves 2. Replacing φ by σaφ we may assume τ = (2, i), for some i. Let a = (3, i). Then σaφ fixes (1, 2) and (2, 3). Replacing φ by σaφ we may assume φ fixes (1, 2) and (2, 3). Continuing in this way, we reduce to the case when φ fixes (1, 2), (2, 3), . . . , and (n − 1, n). As these transpositions generate Sn, φ is then the identity, which is an inner automorphism.

Lemma 23.14. Let σ ∈ Sn be a permutation. If (1) σ has order 2, (2) σ is not a tranposition, and (3) the conjugacy class generated by σ has cardinality n , 2 then n = 6 and σ is a product of three disjoint tranpositions. Proof. As σ has order two it must be a product of k disjoint tranpositions. The number of these is 1 nn − 2 n − 2k + 2 ... . k! 2 2 2 5 For this to be equal to the number of transpositions we must have 1 nn − 2 n − 2k + 2 n ... = , k! 2 2 2 2 that is n n! = 2k(n − 2k)!k! . 2 It is not hard to check that the only solution is k = 3 and n = 6.

Note that if there is an outer automorphism of S6, it must switch transpositions with products of three disjoint transpositions. So the outer automorphism group is no bigger than Z2. The ﬁnal thing is to actually write down an outer automorphism. This is harder than it might ﬁrst appear. Consider the complete graph K5 on 5 vertices. There are six ways to colour the edges two colours, red and blue say, so that we get two 5-cycles. Call these colourings magic. 5 S5 acts on the vertices of K and this induces an action on the six magic colourings. The induced representation is a group homomorphism i: S5 −→ S6, which it is easy to see is injective. One can check that the tranposition (1, 2) is sent to a product of three disjoint tranpositions. But then S6 acts on the left cosets of i(S5) in S6, so that we get a representation

φ: S6 −→ S6, which is an outer automorphism.